this post was submitted on 02 Dec 2024
14 points (100.0% liked)

NotAwfulTech

385 readers
3 users here now

a community for posting cool tech news you don’t want to sneer at

non-awfulness of tech is not required or else we wouldn’t have any posts

founded 1 year ago
MODERATORS
[email protected]
14
Advent of Code 2024 - Historian goes looking for history in all the wrong places (adventofcode.com)
submitted 3 weeks ago by [email protected] to c/[email protected]
54 comments fedilink hide all child comments
 

copy pasting the rules from last year's thread:

Rules: no spoilers.

The other rules are made up aswe go along.

Share code by link to a forge, home page, pastebin (Eric Wastl has one here) or code section in a comment.

you are viewing a single comment's thread
view the rest of the comments
[–] [email protected] 6 points 3 weeks ago* (last edited 3 weeks ago) (1 children)

2-1: I have quickly run out of hecks to give. This is the sort of problem that gives prolog programmers feelings of smug superiority.

spoiler

#include <string>
#include <iostream>
#include <sstream>

int main() {
  int safe = 0;
  std::string s;
  while (std::getline(std::cin, s)) {
    std::istringstream iss(s);
    int a, b, c;
    if (!(iss >> a >> b)) {
      safe++; continue;
    }
    if (a == b || std::abs(a-b) > 3) continue;
    bool increasing = b > a;
    while (iss >> c) {
      if (b == c || std::abs(b-c) > 3) goto structuredprogrammingisfornerds;
      switch (increasing) {
        case false:
          if (c < b) { b = c; continue; }
          goto structuredprogrammingisfornerds;
        case true:
          if(c > b) { b = c; continue; }
          goto structuredprogrammingisfornerds;
      }
    }
    safe++;
    structuredprogrammingisfornerds:;
  }
  std::cout << safe << std::endl;
}

As usual the second part has punished me for my cowboy code, so I'll have to take a different more annoying tack (maybe tomorrow). Or you know I could just double down on the haphazard approach...

[–] [email protected] 6 points 3 weeks ago* (last edited 3 weeks ago) (1 children)

I decided to double down on 2-2, since bad code is one of life's little pleasures. Where we're going we won't need big-oh notation

spoiler

#include <string>
#include <iostream>
#include <sstream>
#include <vector>
#include <iterator>

template <typename It>
bool seemslegit(It begin, It end) {
    if (std::distance(begin, end) == 1) {
      return true;
    }
    int a = *begin++;
    int b = *begin++;
    if (a == b || std::abs(a-b) > 3) return false;;
    bool increasing = b > a;
    while (begin != end) {
      int c = *begin++;
      if (b == c || std::abs(b-c) > 3) return false;;
      switch (increasing) {
        case false:
          if (c < b) { b = c; continue; }
          return false;
        case true:
          if(c > b) { b = c; continue; }
          return false;
      }
    }
    return true;
}

template <typename It>
void debug(It begin, It end) {
  bool legit = seemslegit(begin, end);
  while (begin != end) {
    std::cout << *begin++ << " ";
  }
  //std::cout << ": " << std::boolalpha << legit << std::endl;
}

int main() {
  int safe = 0;
  std::string s;
  while (std::getline(std::cin, s)) {
    std::istringstream iss(s);
    std::vector<int> report((std::istream_iterator<int>(iss)),
                            std::istream_iterator<int>());
    debug(report.begin(), report.end());
    if (seemslegit(report.begin(), report.end())) {
      safe++;
      std::cout << "\n\n";
      continue;
    }
    for (int i = 0; i < report.size(); ++i) {
      auto report2 = report;
      auto it = report2.erase(report2.begin()+i);
      debug(report2.begin(), report2.end());
      if (seemslegit(report2.begin(), report2.end())) {
        safe++;
        break;
      }
    }
    std::cout << "\n\n";
 }
  std::cout << safe << std::endl;
}

CommentaryDoing this "efficiently" should be possible. since you only need ~2-ish look-back you should be able to score reports in O(n) time. One complication is you might get the direction wrong, need to consider that erasing one of the first two elements could change the direction. But that requires thinking, and shoving all the permutations into a function with ungodly amounts of copying does not.

[–] [email protected] 4 points 3 weeks ago (1 children)

re: 2-2yeah that’s what I ended up thinking. Just try the brute force and if it’s too slow, maybe I’ll try be smarter about it.

[–] [email protected] 5 points 3 weeks ago (1 children)

re: 2-2

I was convinced that some of the Perl gods in the subreddit would reveal some forgotten lore that solved this in one line but looks like my brute force method of removing one element at a time was the way to go.

[–] [email protected] 5 points 3 weeks ago* (last edited 3 weeks ago)

I am now less sleep deprived so can say how to do better somewhat sensibly, albeit I cannot completely escape from C++s verbosity:

2-2

  1. Don't worry about the sequences changing direction. Just call the check function both assuming it is increasing and assuming it is decreasing. This is cheap enough because the wrong branch will fail after 3 elements or so.
  2. When encountering an element that fails, you only need to consider removing the previous element, or the current element. If you can get to the next element removing one of those then you can continue on without any real backtracking.

Updated code:

2-2

#include <string>
#include <iostream>
#include <sstream>
#include <vector>
#include <iterator>

bool valid_pair(const std::vector<int> &arr, int i, int j, bool direction) {
  if (i < 0) return true;
  if (j == arr.size()) return true;
  return    !(arr[i] == arr[j])
         && (direction ? arr[i] < arr[j] : arr[j] < arr[i])
         && (std::abs(arr[j]-arr[i]) <= 3);
}

bool valid(const std::vector<int> &arr, bool direction) {
  int checks = 1;
  for (int i = 1; i < arr.size(); ++i) {
    if (valid_pair(arr, i-1, i, direction)) continue;
    if (checks == 0) return false;
    if (   valid_pair(arr, i-2,  i, direction)
        && valid_pair(arr, i,  i+1, direction)) {
      checks -= 1; i += 1;
    } else if (valid_pair(arr, i-1, i+1, direction)) {
      checks -= 1; i += 1;
    } else return false;
  }
  return true;
}

int main() {
  int safe = 0;
  std::string s;
  while (std::getline(std::cin, s)) {
    std::istringstream iss(s);
    std::vector<int> report((std::istream_iterator<int>(iss)),
                            std::istream_iterator<int>());
    safe += (valid(report, true) || valid(report, false));
  }
  std::cout << safe << std::endl;
}